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  • Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 24 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 24 Maths Textbook Solution.

Answers (1)

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Answer:

\frac{1}{\sin (\mathrm{a}-\mathrm{b})} \log \left|\frac{\sin (\mathrm{x}-\mathrm{a})}{\sin (\mathrm{x}-\mathrm{b})}\right|+\mathrm{c}

Given:

\int \frac{1}{(\sin x-a) \sin (x-b)} d x

Hint:

To solve the statement we have to use formula such as sin(A-B).

Solution: 

\frac{1}{\sin (a-b)} \int \frac{\sin (a-b)}{\sin (x-a)(\sin (x-b))} d x

 =\frac{1}{\sin (a-b)} \int \frac{\sin [(x-b)-(x-a)]}{\sin (x-a)(\sin (x-b))} d x

 =\frac{1}{\sin (a-b)} \int \frac{\sin (x-b) \cos (x-a)-\cos (x-b) \sin (x-a)}{\sin (x-a)(\sin (x-b))} d x

 =\frac{1}{\sin (a-b)} \int \frac{\cos (x-a) d x}{\sin (x-a)}-\frac{\cos (x-b)}{\sin (x-b)} \mathrm{d} \mathrm{x}

 =\frac{1}{\sin (a-b)} \log |\sin (x-a)|-\log |\sin (x-b)|+c

 =\frac{1}{\sin (\mathrm{a}-\mathrm{b})} \log \left|\frac{\sin (x-\mathrm{a})}{\sin (x-\mathrm{b})}\right|+\mathrm{c}

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